Answer
$3.634$
Work Step by Step
We will solve the given integral by using integrate-by-parts formula such as: $\int udv=uv-\int v du$
Here, $u=\ln (3x)$ and $dv=x^{2} \ dx \implies v=\dfrac{x^3}{3}$
$\displaystyle \int_{1}^2 x \ln (3x) x^2 \ dx= \ln (3x) (\dfrac{x^3}{3}) -\int_1^2 (\dfrac{x^3}{3}) (\dfrac{1}{x}) \ dx$
or, $= [(\dfrac{x^3}{3}) \ln (3x)- \dfrac{1}{3} \int x^2 \ dx$
or, $=[(\dfrac{x^3}{3}) \ln (3x) - \dfrac{x^3}{9}]_1^2$
or, $=[(\dfrac{2^3}{3}) \ln (6) - \dfrac{2^3}{9}]-[(\dfrac{1^3}{3}) \ln (3) - \dfrac{1^3}{9}]$
or, $\approx 3.634$
